Extending the result ∫∞0((1−2C(x))2+(1−2S(x))2)dx=4π\int_{0}^{\infty} \left( ( 1 – 2C(x))^{2} + (1-2S(x))^{2} \right) \, dx = \frac{4}{\pi}

While generalizing this result, I succeeded in proving that for α>0, β<1 and 1<2α+β<3, we have

0[(xcosttαdt)2+(xsinttαdt)2]dxxβ=π1βΓ(2αβ)Γ(α)csc(πα+π(β1)2).

My question is

  1. Is this a known result?
  2. My ultimate goal is to examine whether the integral
    I(α,β):=0[(xcosttαdt)2+(xsinttαdt)2]2dxxβ
    has closed from or not for general α and β. I know that
    I(12,0)=2π(log41)andI(1,0)=2π33,
    but I know nothing for the other cases. (Here, the former identity in (1) corresponds to the motivating problem linked above.) Is there any other known result concerning this integral?

A further inspection showed that

0((12C(x))2+(12S(x))2)2dx=16π(122πlog(1+2))1.0516193625061961290,

where

C(x)=x0cos(πt22)dtandS(x)=x0sin(πt22)dt

are Fresnel integrals. Indeed, this corresponds to

I(12,12)=π(42π16log(1+2)).

Note that major inverse symbolic calculators do not yield this result.

Answer

Pardon My Progress...

First, the sum of squared integrals inside the brackets, which for convenience we shall denote by f(x;α), may be rewritten as a single double integral with a little algebra and trigonometry:

f(x;α):=(xcosttαdt)2+(xsinttαdt)2=(xcost1tα1dt1)(xcost2tα2dt2)+(xsint1tα1dt1)(xsint2tα2dt2)=xx(cost1tα1cost2tα2)dt1dt2+xx(sint1tα1sint2tα2)dt1dt2=xx(cos(t1)cos(t2)+sin(t1)sin(t2)(t1t2)α)dt1dt2=xxcos(t1t2)(t1t2)αdt1dt2.

Next, we apply a sequence of two two-variable transformations to put the integral into a more tractable form. The first substitution is a simple scaling transformation, (1)(t1,t2)=(xu1,xu2); the second substitution is the more complicated transformation, (2) (u1u2,u1u2)=(w1,w2):

I(α,β)=0dxxβ[(xcosttαdt)2+(xsinttαdt)2]2=0dxxβ[f(x;α)]2=0dxxβ[xxcos(t1t2)(t1t2)αdt1dt2]2=0dxxβ[11cos[x(u1u2)](x2u1u2)αx2du1du2]2=0dxx4(1α)β[11cos[x(u1u2)](u1u2)αdu1du2]2=0dxx4(1α)β[1dw2w211w2dw1wα2cos(xw1)w21+4w2]2=0dxx4(1α)β[dw11+|w1|dw2wα2cos(xw1)w21+4w2]2=0dxx4(1α)β[20dw11+w1dw2wα2cos(xw1)w21+4w2]2=0dxx4(1α)β[0dw11+w1dw2wα2cos(xw1)14w21+w2]2=0dxx4(1α)β[0dw1cos(xw1)1+w1wα2dw214w21+w2]2.

To perform the integration with respect to w2, I appeal to Gradshteyn 3.197(2): under the conditions |arguβ|<π|βu|<1, and 0<(μ)<(λν), we have the result,

uxλ(x+β)ν(xu)μ1dx=uλ(β+u)μ+νB(λμν,μ)×2F1(λ,μ;λμ;βu).

Let μ=1, ν=12, λ=α, u=1+w1, and β=14w21. Then for 12<(α)<1w1>0,

1+w1wα2dw214w21+w2=(w1+1)α14w21+w1+1B(α12,1)2F1(α,1;α1;14w211+w1)=(w1+1)α(w1+22)(22α1)2F1(α,1;α1;w214(1+w1))=(w1+2)(w1+1)α2α12F1(α,1;α1;w214(1+w1)).

The hypergeometric function above quite conveniently reduces to a rational function for the specified combination of parameters: for α1,

2F1(α,1;α1;z)=αz+α2z1(α1)(z+1)2,2F1(α,1;α1;w214(1+w1))=4(w1+1)[α(w1+2)22(w1+1)22](α1)(w1+2)4.

Thus, for 12<(α)<1w1>0,

1+w1wα2dw214w21+w2=(w1+2)(w1+1)α2α12F1(α,1;α1;w214(1+w1))=4(w1+1)1α[α(w1+2)22(w1+1)22](2α1)(α1)(w1+2)3,

and hence:

I(α,β)=0dxx4(1α)β[0dw1cos(xw1)1+w1wα2dw214w21+w2]2=0dxxβ4(1α)[04(w1+1)1α[α(w1+2)22(w1+1)22]cos(xw1)(2α1)(α1)(w1+2)3dw1]2=0dxxβ4(1α)[04(y+1)1α[α(y+2)22(y+1)22]cos(xy)(2α1)(α1)(y+2)3dy]2=16(2α1)2(α1)20dxxβ4(1α)[0(y+1)1α[α(y+2)22(y+1)22]cos(xy)(y+2)3dy]2,=:16(2α1)2(α1)20dxx4(1α)β[0g(y;α)cos(xy)dy]2,

where in the third line we've made the substitution w_{1}=y for the sake of eliminating the dependency on subscripted variables, and where in the last line we've introduced the auxiliary function g{\left(y;\alpha\right)} simply for the sake conveniently denoting the function,

\begin{align}
g{\left(y;\alpha\right)}
&:=\frac{\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]}{(y+2)^3}\\
&=\frac{\alpha\left(y+1\right)^{1-\alpha}}{y+2}-\frac{2\left(y+1\right)^{3-\alpha}}{(y+2)^3}-\frac{2\left(y+1\right)^{1-\alpha}}{(y+2)^3}.\\
\end{align}

Now if we repeat our initial trick of rewriting the square of an integral as a double integral, we can arrive at an expression for \mathcal{I}{\left(\alpha,\beta\right)} as an ordinary triple integral:

\begin{align}
\mathcal{I}{\left(\alpha,\beta\right)}
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]^2\\
&=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]\cdot\left[\int_{0}^{\infty}g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\,\mathrm{d}z\right]}\\
&=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}\right]}\\
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}.\\
\end{align}

Change the order of integration so that the integration over x is first:

\begin{align}
\mathcal{I}{\left(\alpha,\beta\right)}
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\int_{0}^{\infty}\mathrm{d}x\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\int_{0}^{\infty}\mathrm{d}x\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta-4\left(1-\alpha\right)}}.\\
\end{align}

Then the inner integral with respect to x may be evaluated in closed form with an appeal to another result from Gradshteyn. Proposition 3.762(3) of Gradshteyn states that, given a,b\in\mathbb{R} and \mu\in\mathbb{C} such that a>0,~b>0, and 0<\Re{(\mu)}<1, then the following improper integral has the closed form:

\int_{0}^{\infty}x^{\mu-1}\cos{\left(ax\right)}\cos{\left(bx\right)}\,\mathrm{d}x=\frac12\cos{\left(\frac{\mu\pi}{2}\right)}\,\Gamma{\left(\mu\right)}\,\left[\left(a+b\right)^{-\mu}+|a-b|^{-\mu}\right].

Setting (a,b,\mu)\mapsto(y,z,4(1-\alpha)-\beta+1) in the above proposition yields the following corrollary: given y,z\in\mathbb{R} and p\in\mathbb{C} such that y>0,~z>0, and -1<\Re{\left(4(1-\alpha)-\beta\right)}<0, then the following improper integral has the closed form,

\small{\int_{0}^{\infty}\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta+4\alpha-4}}\mathrm{d}x=\frac12\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}\,\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right].}

Thus, we may reduce our integral representation of \mathcal{I}{\left(\alpha,\beta\right)} to a single double integral:

\begin{align}
\mathcal{I}{\left(\alpha,\beta\right)}
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\int_{0}^{\infty}\mathrm{d}x\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\
&=\small{\frac{8\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]}\\
&=:\frac{8\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}}{(2\alpha-1)^2(\alpha-1)^2}\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}.\\
\end{align}


Update:

Now we'll focus on reducing the previously defined function \tilde{\mathcal{I}}{\left(\alpha,\beta\right)}. First of all, by symmetry we can reduce the region of integration to one where the absolute value bars are no longer necessary inside the integrand, which will obviate some of tedium of evaluation:

\begin{align}
\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{y}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}z\int_{0}^{z}\mathrm{d}y\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(z;\alpha\right)}\,g{\left(y;\alpha\right)}\left[\left(z+y\right)^{\beta+4\alpha-5}+|z-y|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+\left(y-z\right)^{\beta+4\alpha-5}\right].\\
\end{align}

Next, we rescale the interval of integration of the inner integral to the unit interval via the substitution z=y\,\omega:

\begin{align}
\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+\left(y-z\right)^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{1}y\,\mathrm{d}\omega\,g{\left(y;\alpha\right)}\,g{\left(y\,\omega;\alpha\right)}\left[\left(y+y\,\omega\right)^{\beta+4\alpha-5}+\left(y-y\,\omega\right)^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{1}y^{\beta+4\alpha-4}\,\mathrm{d}\omega\,g{\left(y;\alpha\right)}\,g{\left(y\,\omega;\alpha\right)}\left[\left(1+\omega\right)^{\beta+4\alpha-5}+\left(1-\omega\right)^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\,y^{\beta+4\alpha-4}\,g{\left(y;\alpha\right)}\int_{0}^{1}\mathrm{d}\omega\,g{\left(y\,\omega;\alpha\right)}\left[\left(1+\omega\right)^{\beta+4\alpha-5}+\left(1-\omega\right)^{\beta+4\alpha-5}\right].\\
\end{align}

At this stage, the inner integral over \omega may be evaluated in terms of the Appell hypergeometric function, F_{1}.

Attribution
Source : Link , Question Author : Sangchul Lee , Answer Author : David H

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